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Suppose that $\mathbf{J}$ is alternating current and its intrinsic frequency is $\omega$, i.e. $\mathbf{J}(\mathbf{r}',t)=\mathbf{J}(\mathbf{r}')e^{-i\omega t}$ , and $\rho$ is charge density. Then my textbook(one Chinese book, no English version, it's called 《电动力学》and written by 郭硕鸿) says they satisfy the law of conservation of charge, so that is $$ i\omega\rho=\nabla\cdot\mathbf{J} \tag{1} $$

I know the continuity equation is $$ -\frac{\partial \rho}{\partial t}=\nabla\cdot\mathbf{J} $$ and it could be derived from the fourth Maxwell's equation. I have tried my best to deduce the equation $(1)$, but I faild.

Here, $\nabla\cdot\mathbf{J}(r',t)=e^{-i\omega t}\nabla\cdot\mathbf{J}(r')$, and from that equation I know $-e^{i\omega t}\frac{\partial \rho}{\partial t}=\nabla\cdot\mathbf{J}(r',t)$, however, I don't know how to do from this.

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  • $\begingroup$ Try $\rho(r',t) = \rho(r') e^{-i \omega t}$. $\endgroup$ – Muphrid Jan 11 '16 at 1:27
  • $\begingroup$ @Muphrid I tried and succeeded, Thank you. And you could write your comment in the answer area, then I will accept your answer. $\endgroup$ – Wang Yun Jan 11 '16 at 1:31
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Like with $J$, it's assumed that $\rho(r',t) = \rho(r') e^{-i\omega t}$--that both the current and charge density oscillate at the same frequency.

In the general case, you can use the orthogonality of complex exponentials to get a frequency-by-frequency equality using the Fourier transform.

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    $\begingroup$ It's really important to note that this works because the equations are linear. $\endgroup$ – DanielSank Jan 11 '16 at 2:06

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